NL, vm 
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Remarks on Several Subjects. 135 
stant) dv=o, and the differential becomes d'T=o0, there- 
fore the maximum or minimum of u will be obtained by 
putting the differential of ‘T equal to nothing. 
It will sometimes be more =n to Be aut (instead 
of y=xt) and then we shall have u=y" T Z 
and T’’ being in general different pie the. values above 
found) hence, we get as above 
Gee f= 
™ (=3 =) 
and the maximum or minimum is found, as before, by put- 
ting du=o, dv=o which gives dT =o. 
It may so happen that the proposed ene without 
any reduction, appear under the form u=x™. T’, v=c", 
; T’,'T”, being functions of y alone without a. In this 
case no reduction will be necessary, because we have im- 
mediately = aT) whose differential gives dT=o, 
when wis amaximum or minimum, The same thing takes 
place, if u=y" T’; v=y". T’; ‘I’, T”, being functions of 
u 
vm 
Me 
if fis: m 
or minimum of « is found, as before, by putting dT =o, 
€ may observe that generally when dT! =o, we shall 
also have d. T?=o0, therefore instead of T, we may take 
any power p of T, positive or negative ; we ma 
neglect any constant factor a by which T is multiplied, 
since d. aT’ =o, gives a. dT =o, whence dI'=o0. 
All the srohbiaes actually solved by professor Fisher, de 
pend on homogeneous functions and can therefore be sol- 
ved by the method just mentioned, putting d'=o ; as we 
thall show by solving a few of his problems. 
Problem 1. Suppose u=*y/yytox, v= 3" y?r, T=3, 
14159, and let it be required to find the value of u when a 
maximum, for a given value of v. 
The substitution of c=yt, gives u=*. y?. /I+tt; v= 
37. y*.t. Comparing with the above forms u=y", T’, > 
=y". T”, we get m=2 n=3, T=1/1 ere, T’=1«t, and 
if we neglect the constant factors in ‘2 = it becomes 
